\subsection{Introduction}
A vector field assigns a vector \(\vb v\) to every position \(\vb x \in \mathbb R^3\).
A scalar field assigns a scalar \(\phi\) to every position.
We generalise this notion to a \textit{tensor field} of rank \(n\), written \(T_{ij\dots k}(\vb x)\), which assigns a rank \(n\) tensor to every point \(\vb x\).
Recall that
\[
	x_i' = R_{ij}x_j \iff x_j = R_{ij}x_i'
\]
Differentiating both sides with respect to \(x_k'\), we get
\[
	\pdv{x_j}{x_k'} = R_{ij} \pdv{x_i'}{x_k'} = R_{ij} \delta_{ik} = R_{kj}
\]
By the chain rule, we then have
\[
	\pdv{x_i'} = \pdv{x_j}{x_i'} \pdv{x_j} = R_{ij} \pdv{x_j}
\]
Informally, we can say that \(\pdv{x_i'}\) transforms like a rank 1 tensor.
\begin{proposition}
	If \(T_{i\dots j}\) is a tensor field of rank \(n\), then
	\[
		\underbrace{\pdv{x_p} \cdots \pdv{x_q}}_{m \text{ terms}} T_{i\dots j}(\vb x)
	\]
	is a tensor field of rank \(n+m\).
\end{proposition}
\begin{proof}
	We check the transformation under a change of basis.
	Let the above expression be \(A_{p\dots q i \dots j}\).
	Then
	\begin{align*}
		A_{p\dots q i\dots j}' & = \pdv{x_p'} \cdots \pdv{x_q'} T_{i\dots j}'(\vb x)                            \\
		                       & = R_{pa}\pdv{x_a} \cdots R_{qb}\pdv{x_b} R_{ic}\dots R_{jd}T_{c\dots d}(\vb x) \\
		                       & = R_{pa}\dots R_{qb}R_{ic}\dots R_{jd} A_{a\dots bc\dots d}
	\end{align*}
\end{proof}
Note that this only works in Cartesian coordinates, since the \(R\) matrices are constant here.
In a general coordinate system, this is not the case, and we cannot move the change of basis matrices outside the derivatives in this case.

\subsection{Differential operators producing tensor fields}
If \(\phi\) is a scalar field, then
\[
	[\grad\phi]_i = \pdv{\phi}{x_i}
\]
Hence \(\grad\phi\) is a rank 1 tensor field, which is a vector field.
If \(\vb v\) is a vector field,
\[
	\div \vb v = \pdv{v_i}{x_i}
\]
which is a rank 0 tensor field since it is a contraction of \(\pdv{\vb v_i}{x_j}\).
Alternatively, from first principles,
\[
	\pdv{v_i'}{x_i'} = R_{ip}\pdv{x_p}R_{iq}v_q = R_{ip}R_{iq}\pdv{v_q}{x_p} = \delta_{pq}\pdv{v_q}{x_p} = \pdv{v_i}{x_i}
\]
hence the divergence of a vector field really is a scalar field.
\[
	[\curl \vb v]_i = \varepsilon_{ijk} \pdv{v_k}{x_j}
\]
From first principles we can show that
\begin{align*}
	\varepsilon_{ijk}' \pdv{v_k'}{x_j'} & = R_{ia}R_{jb}R_{kc}\varepsilon_{abc} R_{jp}\pdv{x_p} R_{kq}v_q       \\
	                                    & = R_{ia} \varepsilon_{abc} R_{jb} R_{jp} R_{kc} R_{kq} \pdv{v_q}{x_p} \\
	                                    & = R_{ia} \varepsilon_{abc} \delta_{bp} \delta_{cq} \pdv{v_q}{x_p}     \\
	                                    & = R_{ia} \varepsilon_{abc} \pdv{v_c}{x_b}
\end{align*}
which is the transformation law for a rank 1 tensor, so the curl of a vector field is a vector field.

\subsection{Divergence theorem with tensor fields}
\begin{proposition}
	For a tensor field \(T_{ij\dots k \dots \ell}(\vb x)\), we have
	\[
		\int_V \pdv{x_k} T_{ij\dots k \dots \ell} \dd{V} = \int_{\partial V} T_{ij\dots k \dots \ell} n_k \dd{S}
	\]
\end{proposition}
\begin{proof}
	Consider the vector field
	\[
		v_k = a_i b_j \dots c_\ell T_{ij\dots k \dots \ell}
	\]
	where the \(a_i, b_j, \dots, c_\ell\) are the components of some constant vectors.
	Applying the divergence theorem to this vector field, we have
	\begin{align*}
		\int_V \pdv{v_k}{x_k} \dd{V}                                         & = \int_{\partial V} v_k n_k \dd{S}                                          \\
		a_i b_j \dots c_\ell \int_V \pdv{x_k} T_{ij\dots k\dots \ell} \dd{V} & = a_i b_j \dots c_\ell \int_{\partial V} T_{ij\dots k\dots \ell} n_k \dd{S}
	\end{align*}
	Since this is true for any choice of vectors \(a_i, b_i, \dots, c_i\), the result follows.
\end{proof}
